how to extract a string from a list of strings from a parameter using bash

Question

I am getting a list of quoted strings back from AWS, and I would like to isolate one of them into a parameter.

the command I am using is:

allelb=$(aws elb describe-load-balancers --query 'LoadBalancerDescriptions[].LoadBalancerName') &&  echo $allelb

And this is the output I am getting:

[ "elb-app-mprest-dev", "elb-core-mprest-dev", "api-vector-k8s-local-0j8ccl", "a2e6a899d111011e897b0067693cf815", "api-clusters-sydney7-mpre-rqae1h" ]

What I want to do is, get only the string with the word 'sydney7' inside a parameter. So I will have a new parameter with this content inside :

ELB=api-clusters-sydney7-mpre-rqae1h

This is waht I got so far:

allelb=$(aws elb describe-load-balancers --query 'LoadBalancerDescriptions[].LoadBalancerName' | tr -d '"' | tr -d ',' | tr -d ']' | tr -d '[') &&  echo $allelb | grep -o sydney7

but the output I get is not good enough-

sydney7

How can I achieve that?

@choroba There will be no more than one from this string since this name is unique — Shahar Hamuzim Rajuan
– Shahar Hamuzim Rajuan, Commented Feb 28, 2018 at 15:08

amphetamachine · Accepted Answer · 2018-02-28 19:48:19Z

3

Since the output from your command is a JSON array, you should probably use a program for parsing and filtering JSON, like jq:

$ jq '.[] | match(".*sydney.*").string' <<< "$allelb"
"api-clusters-sydney7-mpre-rqae1h"

If you want it raw, use the -r flag:

$ jq -r '.[] | match(".*sydney.*").string' <<< "$allelb"
api-clusters-sydney7-mpre-rqae1h

edited Feb 28, 2018 at 19:48

answered Feb 28, 2018 at 15:06

amphetamachine

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1 Comment

Shahar Hamuzim Rajuan Over a year ago

Thanks for the answer, but the assumption is that I can not install jq on my machine, I gave it a vote up :)

ctac_ · Accepted Answer · 2018-02-28 16:02:14Z

1

With cut

cut -d '"' -f10

with grep

grep -o '[^"]*sydney7[^"]*'

answered Feb 28, 2018 at 16:02

ctac_

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